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Difference between revisions of "2010 AMC 8 Problems/Problem 15"

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==Solution==
 
==Solution==
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We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25\%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace all blue gumdrops with brown gumdrops, then <math>35\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
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We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25\%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then <math>15\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
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==Video by MathTalks==
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https://www.youtube.com/watch?v=6hRHZxSieKc
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==See Also==
 
==See Also==
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{{AMC8 box|year=2010|num-b=14|num-a=16}}
 
{{AMC8 box|year=2010|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:55, 22 January 2023

Problem

A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution

We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc



See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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